Every Weakly Compact Set In X Is Norm Bounded

# Every Weakly Compact Set in X is Norm Bounded

 Theorem 1: Let $X$ be a normed linear space and let $K \subset X$. If $K$ is weakly compact then $K$ is norm bounded.
• Proof: Suppose that $K$ is weakly compact.
• Since $X$ is a normed space, for each $k \in K$ by one of the corollaries to Hahn-Banach theorem on the Corollaries to the Hahn-Banach Theorem there exists a $f_k \in X^*$ such that $\| f_k \| = 1$ and $f_k(k) = \| k \|$.
• Since each $f_k \in X^*$ and since $K$ is compact, we have by the Extreme Value theorem that $f_k$ attains a maximum and minimum on $K$, that is, $f_k(K)$ is bounded. So for each $k \in K$:
(1)
\begin{align} \quad \sup_{t \in K} | f_k(t) | = \sup_{t \in K} | \hat{t}(f_k) \| < \infty \end{align}
(2)
\begin{align} \quad \sup_{t \in K} \| \hat{t} \| < \infty \end{align}
• But $\| \hat{t} \| = \| t \|$, and so:
(3)
\begin{align} \quad \sup_{t \in K} \| t \| < \infty \end{align}
• Thus $K$ is norm bounded. $\blacksquare$